MATH 802: ENUMERATIVE COMBINATORICS ASSIGNMENT 2

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1 MATH 80: ENUMERATIVE COMBINATORICS ASSIGNMENT KANNAPPAN SAMPATH Facts Recall that, the Stirling number S(, n of the second ind is defined as the number of partitions of a [] into n non-empty blocs. We recall the following result from class: Lemma 0.1. The following numbers are equal: (1 The number of surjective maps from [] to [n]. ( S(, n where S(, n denotes the Stirling number of the second ind. (3 n i0 ( 1i( n i (n i. Proof. (1 (: Given a partition of [] into n blocs, one may define a surjective map by mapping distinct parts to distinct numbers between 1 and n. There are such surjections. (1 (3: (Principle of Inclusion and Exclusion. Let A i denote the set of maps f : [] [n] such that i / Im(f. Then, it follows that the number of surjective maps [] [n] is given by [n][] \ A i. Applying the principle of inclusion-exclusion, we conclude: [n][] \ A i ( 1 I A I This completes the proof. i1 i1 I [] ( 1 i (n i. i i0 Problems Problem 1. Let S(n, be the number of ways of partitioning an n-set into blocs. Prove that S(n, 1 1, S(n, n 1 1 and S(n, n 1 ( n, Find a formula for S(n, n. Solution. Proof that S(n, 1 1: There is exactly one partition of an n-set which consists of just one bloc, viz, [n]. So, S(n,

2 KANNAPPAN SAMPATH Proof that S(n, n 1 1: Let Σ(n, denote the set of all partitions of [n] into parts (so that Σ(n, S(n,. Note that any element of Σ(n, is of the form A A c for a subset A of [n] such that A A c. Consider the map: ϕ : [n] \ {, [n]} Σ(n, A A A c Then, this map is clearly surjective and is -to-1, indeed ϕ(a ϕ(a c. Thus, Σ(n, [n] \ {, [n]} n whence we have S(n, n 1 1 as required. Proof that S(n, n 1 ( n : Let Σ(n, n 1 denote the set of all partitions of [n] into n 1 parts (so that Σ(n, n 1 S(n, n 1. Note that any element of Σ(n, n 1 is of the form A 1 A An 1 where A i is non-empty for every i. For a fixed partition of the above form, let a i A i. Then, we have that a i 1 and n 1 a i n. i1 If there are atleast two indices, say i, j such that a i, a j, then, we have that n 1 n a i + + (n 1 n + 1 i1 a contradiction. On the other hand, if a i 1 for all i, then, we get the absurd equation n n 1. Thus, there is precisely one index i such that a i and a j 1 for all j i. We may let i n 1 without loss of generality. (This is an elaboration of the Pigeonhole principle. The above consideration shows that an element of Σ(n, n 1 is determined by the -set A n 1. Conversely, every -subset T of [n] determines a partition of [n] into n 1 blocs: [n] a/ T{a} T. This shows that S(n, n 1 ( n (the number of -subsets of [n]. Formula for S(n, n : In this part, we argue as in the previous part, but somewhat more systematically. Let c denote the number of blocs with cardinality c in a given partition of [n] into n blocs. Then, we have the equation: (n c c + c c c n c (c 1 c. This equation shows that if c > 0, then c 3. Further it follows that, 3 0 or 0, 3 1. The number of partitions of [n] into n blocs consisting of two blocs of size (and other blocs singleton is ( 1 n ( n. The number of partitions of [n] into n blocs consisting of one bloc of size 3 (and other blocs singleton is ( n 3.

3 MATH 80: ENUMERATIVE COMBINATORICS ASSIGNMENT 3 Based on the above analysis, it follows that S(n, n + 1 ( ( n n n(n 1(n (3n This completes this problem. Remar 0.. Another way to approach the above problem would be through recurrence established for Stirling numbers S(n, of the second ind. Recall from class that S(n, S(n 1, 1 + S(n 1, Set a n S(n, n ; we now that a 3 S(3, 1 1. On taing n in the above recurrence relation, we have 1 a n a n 1 + (n using the previous calculation. Rearranging the above equality and summing from n 4 to gives us: a 1 (a n a n 1 n4 (n 1(n n4 Maing a change of variable in the above sum: n n (n n 3 + n n Using the nown formulae: 1 ( ( ( 1 ( ( 1( ( ( 1( ( ( 1 a ( 1( (3 5 4 This also proves the expression we computed for S(n, n above. n0 1 Problem. Let f n and g n be two sequences with exponential generating functions f n t n g n t n F (t, G(t. Show that the following are equivalent: (a g n 0 f ; n0

4 4 KANNAPPAN SAMPATH (b G(t F (te t. Proof. Since an exponential generating function uniquely determines the sequence and conversely, the equivalence is immediate in view of the following calculation: g n f G(t G(t Interchanging the order of summation: G(t 0 n0 n0 0 t n t n 0 0 t! f f!(n! f n Maing the change of variable n u in the inner sum: t G(t! f t u u! This finishes the proof. 0 G(t F (te t u0 t n (n! Problem 3. Let d n be the number of derangements of an n-element set (that is, the number of elements of S n without any fixed points. Show that d n nd n 1 + ( 1 n Deduce that d n is odd if and only if n is even. Proof. We have shown in class that d n satisfies the following recurrence: d n (n 1(d n 1 + d n, for n > 1 d 0 1, d 1 0. Substracting nd n 1, the relation becomes: d n nd n 1 (d n 1 (n 1d n Setting c n d n nd n 1, we see that c n c n 1 with c 1 1. Thus, it follows immediately that c n ( 1 n which proves the relation required. Now, the final parity argument is easy: if n is even, then, d n ( 1 n 1 mod, so d n is odd. On the other hand, if n is odd, then, d n is even since n 1 is even and we have just shown that d n 1 is odd; using the fact that d n nd n 1 1, we have that d n is even. This finishes the problem. Problem 4. The Bernoulli numbers b n are defined by the recurrence b 0 1 and + 1 b 0 0

5 MATH 80: ENUMERATIVE COMBINATORICS ASSIGNMENT 5 for n 1. Prove that f(t : n0 b n tn t e t 1. Proof. From the given recurrence, we have b n 1 n + 1 n 1 ( n b for n > 0. Now, we compute the function f(t as follows: Put m n 1: f(t n b n t n n1 n1 m0 t n (n + 1! n 1 ( n b n 1 t n 1!(n + 1! b 0 t m+1 m 0 Interchanging the order of summation, we have: Put v m + to get: This finishes the proof m 0 v 0 t 1 b! 1!(m +! b t m+1!(m +! b t v+ 1 b!v! v f(t 1 f(t(et t 1 t f(t t e t 1. t v v! Problem 5. In the previous question, show that f(t + t is an even function of t (that is, it is invariant under the map t t. Deduce that b n 0 for all odd n 3.

6 6 KANNAPPAN SAMPATH Proof. To prove the first part, we simply compute: f( t + t t e t 1 t tet t(e t 1 (e t 1 tet + t (e t 1 t ( e t + 1 e t 1 t ( e t 1 + e t 1 t ( 1 + e t 1 t + t e t 1 f(t + t Now, plugging our power series for f(t, we get: ( 1 n b n t n n0 + t n0 b n t n Comparing the coefficient of t n on both sides implies that b 1 1/ and that b n 0 for all odd n 3. Problem 6. Let s(n, and S(n, denote the Stirling numbers of the first and second ind respectively. Show that S(n, s(n,. Deduce that the n-th Bell number satisfies B n. Proof. By definition, we have the following descriptions: S(n, : the number of (unordered partitions of [n] into blocs. s(n, : the number of elements in S n that can be written as a product of disjoint cycles. B n : the number of (unordered partitions of [n]. Based on these descriptions, it is easy to deduce the claim about Bell numbers: indeed, we have B n S(n, 1 s(n,. 1 Thus, to finish the problem, we must show that S(n, s(n,. We show that there is an injection from the set Σ(n, of partitions of [n] into blocs into the set of elements in S n that can be written as a product of disjoint cycles. Indeed, given an element σ σ[1] σ[] Σ(n,, we may construct a permutation + t

7 MATH 80: ENUMERATIVE COMBINATORICS ASSIGNMENT 7 σ as follows: σ (σ σ 1r1... (σ 1... σ r where σ j1 < < σ jrj and σ[j] {σ j1,..., σ jrj }. This map is clearly an injection and this shows that S(n, s(n,. Problem 7. With B n denoting the n-th Bell number (that is, the number of partitions of an n-element set, show that n n B n ( 1 j.! j! 1 j0 Proof. We begin by noting B n S(n, 1 where S(n, is the Stirling number of the second ind. Thus (for n > 0: 1 1 1! 1 1 i0 i0 Substituting j i, it follows that: 1 j1 ( 1 i ( i ( i n. ( 1 i 1 i!( i! ( in. ( 1 j 1 ( j!j! jn Interchanging the order of summation, it follows that: ( 1 j 1 ( j!j! jn j1 j j1 j n j! ( 1 j 1 ( j! j Maing a change of variable in the above sum, it follows that: This finishes the proof. j1 j n n j ( 1 1 j!! 1 Problem 8. A permutation σ of the symmetric group S n is called an involution if σ 1. Let s(n denote the number of involutions of S n. Defining s(0 1, prove that s(nt n ( exp t + t. n0

8 8 KANNAPPAN SAMPATH Proof. We first establish a recurrence relation for s(n. Let σ S n such that σ 1. If σ(n n, then, σ is an involution in S n 1. Otherwise, (nσ(nσ is an involution on n symbols. This gives us the following recurrence: s(n s(n 1 + (n 1s(n. Let S(t s(nt n n0. We now do the following: s(nt n (1 + ts(t (1 + t n0 s(nt n s(nt n+1 + n0 n0 s(nt n s(n 1t n + (n 1! n0 n1 s(nt n ns(n 1t n + n0 n1 s(n + ns(n t n n1 s(n t n n1 ns(nt n n ns(nt n 1 n1 S (t This proves that S(t exp(t + t / since S(0 1. Problem 9. Prove that ( ( n m + n i ( 1 i i i i0 ( m if m and 0 otherwise. [Hint: if you have m red cards and n blue cards, count the number of -element subsets consisting solely of red cards.] Proof. Continuing the notation in Hint, let A denote the set of all -subsets of m + n cards. Index the blue cards from 1 to n. For each i [n], let A i denote the set of -subsets of m + n cards which consists of the ith blue card. The number of -subsets of m + n cards which consists only of red cards is then given by: n A \ {( m A i, if m 0, otherwise. i1

9 MATH 80: ENUMERATIVE COMBINATORICS ASSIGNMENT 9 Applying the Principle of Inclusion-Exclusion, we see: m (1 A \ A i ( 1 I A I i1 I [n] where A I i I A i (with the obvious convention that A A. Clearly, A I is just the set of -subsets of m + n cards which contains all the blue cards whose indices are in I. If I i, then, this number is just ( m+n i i. Notice that this number is independent of the subset I (depends only on I. Now, continuing from (1: m A \ A i ( 1 I A I This completes the proof. i1 I [n] ( ( n m + n i ( 1 i. i i i0 Problem 10. Let A n denote the number of maps f : [n] [n] with the property that if i lies in the image of f, then, so does [i]. Prove that Deduce that A n 1 A n A n 0 n +1 In this problem, we have (by convention, for the scholars of the empty set that A 0 1. Proof. We reinterpret the number A n. Firstly, a function f : [n] [n] that satisfies the hypothesis of the problem must surject onto [] for some 1 n and conversely; now, a surjective function f : [n] [] determines an ordered partition of [n] into blocs and conversely, given a partition P : S 1 S S [n] of [n] into blocs, we may construct f P : [n] [] by f P (j l if j S l. Thus, A n is the number of ordered partition of [n] into non-empty blocs (for n > 0; let A n denote the set of ordered partitions of [n]. Given an ordered partition P of [n], we write P(1 for the first bloc of the given partition. Then, we have the following count (assuming that n > 0: A n 1 P A n 1 S [n] P[1]S

10 10 KANNAPPAN SAMPATH The inner sum counts the number of ordered partitions with first bloc S which is the same as the number of ordered partitions of [n S ]: S [n] 1 A n S A n This finishes the proof of the first equality. Now to prove the second equality, we proceed by induction on n. For n 0, the result holds. Indeed, the sum defining A 0 is a geometric series with initial term 1/ and common ratio 1/ whence its sum is 1 (by convention, as always, Since the recursion holds for all n 1, we plug this explicit formula into the recursion and get that: A n A n 1 1 l0 ( n l0 1 l n l+1 Interchanging the order of summation: 1 l+1 l n We eep the term l 0 out (and implement our convention that 0 0 1: 1 + l n l+1 l l1 1 + l1 1 + l1 l n l+1 1 (( l n 1 (l + 1 n l+1 l1 Maing a change of variable in the first sum: 1 + l n l l n l+1 This completes the problem. l1 l l n l+1 l1 l n l+1